CHEM 1251 Fall 2024 Week 6 Problem Session PDF

# CHEM 1251 Fall 2024 - Week 6 Problem Session ## 1. Solar Cell Efficiency Of the sunlight that reaches Earth, roughly 7% lies in the ultraviolet region of the electromagnetic spectrum, 46% in the visible region, and 47% in the infrared region. Historically, solar cells only used visible waveleng...

# CHEM 1251 Fall 2024 - Week 6 Problem Session ## 1. Solar Cell Efficiency Of the sunlight that reaches Earth, roughly 7% lies in the ultraviolet region of the electromagnetic spectrum, 46% in the visible region, and 47% in the infrared region. Historically, solar cells only used visible wavelengths and the shortest of the IR wavelengths, but recently new designs have made use of longer IR wavelengths possible - increasing the efficiency of solar panels. The longest wavelength reported to be harvested by a solar cell is 1750 nm. **(a) Determine the frequency of this radiation.** * $c = \lambda \nu \implies \nu = \frac{c}{\lambda}$ * $\nu = \frac{3.00 \times 10^8 \text{ m/s}}{1.750 \times 10^{-6} \text{ m}} = 1.71 \times 10^{14} \text{ Hz}$ **(b) Determine the energy per photon of this radiation.** * $E = h\nu \implies E = \frac{hc}{\lambda}$ * $E = \frac{(6.63 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{1.75 \times 10^{-6} \text{ m}}$ * $E = 1.14 \times 10^{-19} \text{ J/photon}$ **(c) Determine the energy per mole of photons of this radiation.** * $E_{mol} = (1.14 \times 10^{-19} \text{ J/photon})(6.022 \times 10^{23} \text{ photons/mol})$ * $E_{mol} = 6.87 \times 10^4 \text{ J/mol}$ ## 2. Atomic Properties Determine which element has the greatest... **(a) Zeff.** * Ge: $Zeff = 32 - 28 = 4$ * Ca: $Zeff = 20 - 18 = 2$ * Se: $Zeff = 34 - 28 = 6$ **Answer: Se** **(b) Size.** * Atomic size increases as you move down and to the left on the periodic table. * Atomic size decreases as you move up and to the right. * $Zeff$ also affects size, with a higher $Zeff$ leading to a smaller radius due to the greater pull of the nucleus. **Answer: N** **(c) Average atomic mass.** **Answer: I** **(d) First ionization energy.** * First ionization energy increases as you move up and to the right. **Answer: He** **(e) First ionization energy.** * **Answer: S** * It is possible that N is the answer as well, as the trend for ionization energy is not strictly followed, but the general trend shows that S should have a greater ionization energy than Br. ## 3. Aluminum Ionization **(a) The first ionization energy of aluminum is 578 kJ/mol. How much energy, in J, is required to ionize one atom of Al?** * $578 \text{ kJ/mol} \times \frac{10^3 \text{ J}}{1 \text{ kJ}} \times \frac{1 \text{ mol}}{6.022 \times 10^{23} \text{ atoms}} = 9.60 \times 10^{-19} \text{ J}$ **(b) From which subshell is the electron ionized? Use the condensed electron configuration to help you describe this, and give the quantum numbers for this electron.** * Aluminum has a condensed electron configuration of $[Ne]3s^23p^1$ * Therefore, the electron that is ionized is in the 3p subshell. * The quantum numbers for that electron are: n=3, l=1, ml=0, s=+1/2 **(c) If it takes one photon with energy equal to or greater than the ionization energy to ionize an atom, what is the longest wavelength of light that would ionize aluminum?** * $E_{photon} = h\nu = \frac{hc}{\lambda}$ * $\lambda = \frac{hc}{E_{photon}} = \frac{(6.63 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{9.60 \times 10^{-19} \text{ J}} = 2.07 \times 10^{-7} \text{ m}$ **(d) It would take a photon of {higher | lower } wavelength to ionize silicon compared to aluminum. Explain:** * It would take a photon of **higher** wavelength to ionize silicon. * Ionization energy increases to the right on the periodic table, meaning silicon has a greater ionization energy than aluminum. * Because the energy of a photon is inversely proportional to its wavelength, a higher ionization energy means silicon would require a photon with lower wavelength (higher energy) to ionize. ## 4. Atomic Radius Ranking Arrange these atoms in order of increasing radius: * Ca, Rb, S, Si, Ge, F **Answer:** F, S, Si, Ge, Ca, Rb ## 5. Atomic and Ionic Radius Ranking Arrange these atoms or ions in order of increasing radius: * S^2-, Ca²⁺, Cl^-, K⁺, Ar * **Key points:** * Cations (positively charged ions) are smaller than their neutral atoms because they have lost an electron, making the remaining electrons more strongly attracted to the nucleus. * Anions (negatively charged ions) are larger than their neutral atoms because they have gained an electron, increasing the electron-electron repulsion, and making the electron cloud larger. **Answer:** Ca²⁺, K⁺, Ar, Cl^-, S^2- ## 6. Atomic and Ionic Radius Ranking Arrange these atoms or ions in order of increasing radius: * Br^-, Br, I^-, I, Kr * **Key points:** * Atomic radius increases going down and to the left on the periodic table. * Ions have different radii than neutral atoms based on the number of electrons they gain or lose. **Answer:** Kr, Br, Br^-, I, I^- ## 7. Copper Sphere Volume A pure copper sphere contains 8.62 x 10²⁷ electrons. What is the volume of the sphere in m³? Assume the sphere is made up of neutral copper atoms. The density of copper is 8.96 g/mL. * **Step 1:** Calculate the number of copper atoms. * Each copper atom has 29 electrons. * Number of copper atoms = (8.62 x 10²⁷ electrons) / (29 electrons/atom) = 2.97 x 10²⁶ atoms. * **Step 2:** Calculate the mass of the copper sphere. * Mass of copper atoms = (2.97 x 10²⁶ atoms) x (63.55 g/mol) / (6.022 x 10²³ atoms/mol) = 7.76 g. * **Step 3:** Calculate the volume of the copper sphere. * Density = Mass / Volume * Volume = Mass / Density = 7.76 g / (8.96 g/mL) = 0.866 mL * Convert mL to m³: 0.866 mL x (1 m³ / 10⁶ mL) = 8.66 x 10^-7 m³ **Answer:** The volume of the copper sphere is 8.66 x 10^-7 m³.